Particle dynamics
• Second-order motion • Particle system • Forces • Constraints • Second order motion analysis (advanced)
Particle system • Particles are objects that have mass, position, and velocity, but without spatial extent • Particles are the easiest objects to simulate but they can be made to exhibit a wide range of objects
A Newtonian particle • First order motion is sufficient, if • a particle state only contains position • no inertia • particles are extremely light • Most likely particles have inertia and are affected by gravity and other forces • This puts us in the realm of second order motion
Second-order ODE What is the differential equation that describes the behavior of a mass point? f = m a What does f depend on? x ( t ) = f ( x ( t ) , ˙ x ( t )) ¨ m
Second-order ODE x ( t ) = f ( x ( t ) , ˙ x ( t )) ¨ = f ( x , ˙ x ) m This is not a first oder ODE because it has second derivatives Add a new variable, v ( t ), to get a pair of coupled first order equations { ˙ x = v v = f /m ˙
Phase space ⎡ ⎤ x 1 x 2 Concatenate position and ⎢ ⎥ � x ⎢ ⎥ � x 3 ⎢ ⎥ velocity to form a 6-vector: = ⎢ ⎥ v v 1 ⎢ ⎥ position in phase space ⎢ ⎥ v 2 ⎣ ⎦ v 3 First order differential ˙ v � � � x x = f ( ) = equation: velocity in f ˙ v v m the phase space
Quiz A mass point attached to a spring obeys Hooke’s Law: f = − K ( x − ¯ x ) What is the ODE that describes this motion?
Integrate second-order ODE Express a second-order motion in two first-order ODEs, ✓ ˙ ◆ ✓ x ◆ ✓ ◆ ✓ ◆ x 0 1 0 = + ˙ ( K /m )¯ − K /m v 0 v x Integrate both position and velocity via explicit Euler ✓ ˙ ✓ x 1 ✓ x 0 ◆ ◆ ◆ x 0 = + h ˙ v 1 v 0 v 0 ✓ x 0 ◆ ✓ ◆ v 0 = + h K /m (¯ x − x 0 ) v 0
Quiz • Integrate the same ODE using midpoint method. ✓ ˙ ◆ ✓ x ◆ ✓ ◆ ✓ ◆ x 0 1 0 = + ˙ ( K /m )¯ − K /m v 0 v x
• Second-order motion • Particle system • Forces • Constraints • Second order motion analysis (advanced)
Particle structure Particle position x a point in the phase space v velocity f force accumulator mass m
Solver interface solver interface system solver 6 GetDim particle x x v Get/Set State v f v m Deriv Eval f m
Particle system structure system x 1 x 2 x n v n v 1 v 2 particles ... f n f 1 f 2 m n m 1 m 2 n time
Particle system structure solver system solver interface 6n GetDim particles x 1 x 2 x n n time Get/Set State . . . v 1 v 2 v n v 1 v 2 v n Deriv Eval f 1 f 2 f n . . . m 1 m 2 m n
Deriv Eval Clear forces: loop over particles, zero force accumulator Calculate forces: sum all forces into accumulator Gather: loop over particles, copy v and f /m into destination array
• Second-order motion • Particle system • Forces • Constraints • Second order motion analysis (advanced)
Forces • Constant gravity • Position dependent force fields, springs • Velocity dependent drag
Particle systems with forces system x 1 x 2 x n v n v 1 v 2 particles ... f n f 1 f 2 m n m 1 m 2 n time ... forces F 1 F 2 F m
Force structure • Unlike particles, forces are heterogeneous (type-dependent) • Each force object “knows” • which particles it influences • how much contribution it adds to the force accumulator
Particle systems with forces system x 1 x 2 x n v n v 1 v 2 particles ... f n f 1 f 2 m n m 1 m 2 n time ... forces F 1 F 2 F m
Gravity x 1 x 2 x n Unary force: f = m G v 1 v 2 v n f 1 f 2 f n m 1 m 2 m n F Exerting a constant p force on each particle . . . sys apply_fun particle system G p->f += p->m*F->G
Viscous drag At very low speeds for small x 1 x 2 x n v 1 v 2 v n particles, air resistance is f 1 f 2 f n m 1 m 2 m n approximately: F p f drag = − k drag v . . . sys apply_fun particle system k p->f += p->v*F->k
Attraction Act on any or all pairs of particles, depending on their positions l f p = − k m p m q l x p | l | 2 | l | x q f q = − f p l = x p − x q
Attraction x p x q v p v q l f p = − k m p m q f p f q m p m q | l | 2 | l | F p sys apply_fun particle system k
Damped spring x p � � ˙ l · l l f p = − k s ( | l | − r ) + k d | l | | l | r f q = − f p | l | l = x p − x q x q
Damped spring x p x q v p v q � � ˙ l · l l f p f q f p = − k s ( | l | − r ) + k d | l | | l | m p m q F r p k d sys apply_fun particle system k s
Quiz For an ideal spring, what is the force it applies to two particles, p and q, attached to it. Write down the pseudo code for its “apply_fun”. x p x q v p v q f p f q m p m q F r p sys apply_fun particle system k s
Deriv Eval 1. Clear force accumulators x 1 x 2 x n v n v 1 v 2 2. Invoke apply_force ... f n f 1 f 2 functions m n m 1 m 2 ... F 1 F 2 F m 3. Return derivatives to solver � ˙ � v � � x = f ˙ v m
ODE solver Euler’s method: x ( t 0 + h ) = x ( t 0 ) + hf ( x , t ) x t +1 = x t + h ˙ x t v t +1 = v t + h ˙ v t
Euler step solver interface system solver x t +1 = x t + h ˙ x t GetDim 3. particles v t +1 = v t + h ˙ v t 4. 2. x 1 x 2 x n 5. Advance time Get/Set State . . . v 1 v 2 v n time 1. v 1 v 2 v n Deriv Eval Deriv Eval f 1 f 2 f n . . . m 1 m 2 m n
Quiz solver interface system solver GetDim Get/Set State Deriv Eval How to modify the algorithm to use midpoint method?
Example: freefall motion • Solution is v ( t ) = v 0 + a 0 t x ( t ) = x 0 + v 0 t + 1 2 a 0 t 2 • v ( t ) only needs 1st order accuracy, but x ( t ) demands 2nd order accuracy
Quiz • Let particle p start at position x 0 with velocity v 0 , what is the state of p after two time steps ( h ) using the midpoint method? Assume that gravity is the only force present in the scene.
• Second-order motion • Particle system • Forces • Constraints • Second order motion analysis (advanced)
Particle Interaction • We will revisit collision when we talk about rigid body simulation • For now, just simple point-plane collisions
Collision detection Particle is on the legal side if ( x − p ) · N ≥ 0 Particle is within of the wall if � x N ( x − p ) · N < � v Particle is heading in if p v · N < 0
Collision response Normal and tangential components x v T v N = ( N · v ) N N v N v T = v − v N v
Collision response Before After collision collision v ′ − k r v N v T v T v N v v ′ = v T − k r v N coefficient of restitution: 0 ≤ k r < 1
Contact Conditions for resting contact: 1. particle is on the collision surface 2. zero normal velocity If a particle is pushed into the contact plane a contact force f c is exerted to cancel the normal component of f N x v p f f N f T
• Second-order motion • Particle system • Forces • Constraints • Second order motion analysis (advanced)
Linear analysis • Linearly approximate acceleration � � � d x x x = f ( , t ) = A + a 0 v v v dt • Split up analysis into different cases • constant acceleration • linear acceleration � � � d x 0 I x + a 0 = v − K − D v dt
Constant acceleration • Solution is v ( t ) = v 0 + a 0 t x ( t ) = x 0 + v 0 t + 1 2 a 0 t 2 • v ( t ) only needs 1st order accuracy, but x ( t ) demands 2nd order accuracy
Linear acceleration • When K (or D) dominates ODE, what type of motion does it correspond to? � � � � � � � � d x 0 I x x = A = v − K − D v v dt • Need to compute the eigenvalues of A
Linear acceleration � � u 1 Assume is an eigenvalue of A , is the α u 2 corresponding eigenvector � � � � � � 0 I u 1 u 1 = α − K − D u 2 u 2 � � u The eigenvector of A has the form α u Assuming D is linear combination of K and I (Rayleigh damping) That means K and D have the same eigenvectors
Linear acceleration � � u For any u , if is an eigenvector of A, the following α u must be true � � � � � � 0 I u u = α − K − D α u α u Now assume u is an eigenvector for both K and D − λ k u − αλ d u = α 2 u � α = − 1 (1 2 λ d ) 2 − λ k 2 λ d ±
Eigenvalue approximation • If D dominates α ≈ − λ d , 0 • exponential decay • If K dominates √ � α ≈ ± λ k − 1 • oscillation
Analysis • Constant acceleration (e.g. gravity) • demands 2nd order accuracy for position • Position dependence (e.g. spring force) • demands stability, oscillatory motion • looks at imaginary axis • Velocity dependence (e.g. damping) • demands stability, exponential decay • looks at negative real axis
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